Let $f(x) = -4x^{2}-8x+1$. Where does this function intersect the x-axis (i.e. what are the roots or zeroes of $f(x)$ )?
Solution: The function intersects the x-axis when $f(x) = 0$ , so you need to solve the equation: $-4x^{2}-8x+1 = 0$ Use the quadratic formula to solve $ax^2 + bx + c = 0$ $x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ $a = -4, b = -8, c = 1$ $ x = \dfrac{+ 8 \pm \sqrt{(-8)^{2} - 4 \cdot -4 \cdot 1}}{2 \cdot -4}$ $ x = \dfrac{8 \pm \sqrt{80}}{-8}$ $ x = \dfrac{8 \pm 4\sqrt{5}}{-8}$ $x =\dfrac{2 \pm \sqrt{5}}{-2}$